Additional: Concepts of Current Density and Conductivity

Electric Current and Current Density ($ \vec{J} = I/\vec{A} $)

We previously defined electric current ($I$) as the total charge flowing per unit time through a conductor. While electric current is a fundamental scalar quantity, it doesn't fully describe how the current is distributed across the cross-section of the conductor. For a more detailed and vector description of charge flow, we introduce the concept of current density.

---

Electric Current ($I$) - A Scalar Quantity

As a quick recap, electric current ($I$) is defined as the time rate of flow of electric charge ($Q$):

$ I = \frac{dQ}{dt} $

Its unit is the Ampere (A), which is 1 C/s. Current is treated as a scalar quantity in circuit analysis, although we often indicate its direction of flow in diagrams. However, the current itself is not a vector because it does not follow vector addition rules (e.g., currents adding at a junction follow scalar addition, not vector addition).

---

Current Density ($\vec{J}$) - A Vector Quantity

Current density ($\vec{J}$) is defined as the electric current flowing per unit cross-sectional area of the conductor. It is a vector quantity. Its direction at any point is the direction of the drift velocity of the positive charge carriers at that point. In the case of electron flow in metals, the direction of current density is opposite to the direction of electron drift velocity, but in the same direction as the conventional current.

For a conductor with uniform cross-sectional area $A$ and uniform current $I$ flowing perpendicular to the area, the magnitude of the current density is given by:

$ J = \frac{I}{A} $

The unit of current density is Ampere per square meter ($A/m^2$).

---

General Definition of Current and Current Density

In a more general case, where the current flow may not be uniform or perpendicular to the cross-sectional surface, the current $I$ flowing through a surface element $d\vec{A}$ is given by the dot product of the current density vector $\vec{J}$ and the area vector $d\vec{A}$ (where $d\vec{A}$ is a vector perpendicular to the surface element, with magnitude equal to the area of the element).

The current through a small area $d\vec{A}$ is $dI = \vec{J} \cdot d\vec{A}$.

The total current $I$ flowing through a surface S is the integral of the current density over that surface:

$ I = \int_S \vec{J} \cdot d\vec{A} $

The direction of $d\vec{A}$ is usually taken to be along the direction of $\vec{J}$ for calculating the total current flow through a cross-section.

---

Relation between Current Density and Drift Velocity

We previously related the current $I$ to the drift velocity $v_d$ of charge carriers: $I = n|q|A|v_d|$, where $n$ is the number density of charge carriers, $q$ is the charge of each carrier, and $A$ is the cross-sectional area.

If we consider a small volume element $dV = A \, dx$ and the carriers drift a distance $dx$ in time $dt$ with drift velocity $v_d = dx/dt$, the charge contained in this volume element is $dQ = nq \, dV = nq A \, dx$.

The current is $I = dQ/dt = nq A (dx/dt) = nq A v_d$.

Now, current density is $J = I/A$.

$ J = \frac{nq A v_d}{A} = n q v_d $

In vector form, the current density $\vec{J}$ is in the direction of positive charge flow. The drift velocity $\vec{v}_d$ is the average velocity of the charge carriers.

$ \vec{J} = n q \vec{v}_d $

Here, $n$ is the number density of charge carriers, and $q$ is the charge of each carrier.

For electrons in metals, $q = -e$ (where $e$ is the elementary charge magnitude). The drift velocity $\vec{v}_d$ is opposite to the electric field $\vec{E}$ (and thus opposite to the conventional current direction). The current density $\vec{J}$ is in the direction of the electric field (conventional current direction).

$ \vec{J} = n (-e) \vec{v}_d $

(Note: $\vec{v}_d$ here represents the actual velocity vector of electrons, which is opposite to $\vec{J}$). If $\vec{v}_d$ is taken as the magnitude of the average electron velocity, then $J = n e v_d$. To maintain vector consistency, using $\vec{J} = n q \vec{v}_d$ is more general.

---

Answer:

Conductivity ($ \sigma = 1/\rho $) and Mobility

While resistance and resistivity quantify how much a material opposes current flow, conductivity quantifies how easily a material conducts current. It is the inverse of resistivity. Mobility describes the ease with which charge carriers move through the material under an electric field.

---

Electrical Conductivity ($\sigma$)

Electrical conductivity ($\sigma$) is defined as the reciprocal of electrical resistivity ($\rho$).

$ \sigma = \frac{1}{\rho} $

A material with high conductivity has low resistivity, and vice versa. Good conductors have high conductivity, while insulators have very low conductivity.

The SI unit of resistivity is ohm-meter ($\Omega \cdot m$). Therefore, the SI unit of conductivity is the reciprocal of ohm-meter, which is $( \Omega \cdot m )^{-1}$. This unit is also called Siemens per meter (S/m), where Siemens (S) is the unit of conductance (reciprocal of resistance, $1/\Omega$).

From our microscopic analysis of resistivity $\rho = \frac{m}{ne^2\tau}$ (for electrons), the conductivity is given by:

$ \sigma = \frac{1}{\rho} = \frac{ne^2\tau}{m} $

This formula shows that conductivity depends on the number density of charge carriers ($n$), the charge of the carriers ($e$), the relaxation time ($\tau$), and the mass of the carriers ($m$).

---

Electrical Mobility ($\mu$)

Electrical mobility ($\mu$) of a charge carrier is defined as the magnitude of the average drift velocity ($v_d$) acquired by the charge carrier per unit electric field strength ($E$).

$ \mu = \frac{|v_d|}{|E|} $

It quantifies how responsive the charge carriers are to the electric field. A higher mobility means that charge carriers can move faster on average for a given electric field.

The SI unit of drift velocity is m/s, and the unit of electric field is V/m. Thus, the SI unit of mobility is $(m/s) / (V/m) = m^2/(Vs)$.

From the expression for drift velocity magnitude, $v_d = \frac{|q|E}{m}\tau$ (where $|q|$ is the magnitude of the charge), we have:

$ \mu = \frac{(|q|E/m)\tau}{E} = \frac{|q|\tau}{m} $

For electrons, $|q|=e$, so $\mu = \frac{e\tau}{m}$.

Mobility is directly proportional to the relaxation time $\tau$ and inversely proportional to the mass of the charge carrier. Lighter particles with longer times between collisions are more mobile.

---

Relation between Conductivity and Mobility

We have the conductivity formula $\sigma = \frac{nq^2\tau}{m}$ (general case for charge $q$) and mobility formula $\mu = \frac{|q|\tau}{m}$.

We can rewrite the conductivity formula as:

$ \sigma = n |q| \left( \frac{|q|\tau}{m} \right) $

Substituting the expression for mobility $\mu = \frac{|q|\tau}{m}$:

$ \sigma = n |q| \mu $

This is a fundamental relationship between conductivity, number density of charge carriers, the magnitude of their charge, and their mobility.

For metals, where the charge carriers are electrons (charge $-e$, magnitude $|-e|=e$):

$ \sigma = n e \mu $

This equation highlights that high conductivity can result from either a high number density of charge carriers ($n$) or high mobility ($\mu$), or both. Metals have very high $n$ and moderately high $\mu$. Semiconductors have lower $n$ than metals but $n$ is temperature-dependent, and mobility is also temperature-dependent. Insulators have extremely low $n$.

---

Answer:

Vector Form of Ohm's Law ($ \vec{J} = \sigma \vec{E} $)

While Ohm's Law is commonly stated as $V = IR$ (a relationship between scalar quantities $V, I, R$), it also has a more fundamental form in terms of the electric field ($\vec{E}$) and current density ($\vec{J}$). This is the vector form of Ohm's Law.

---

Statement of Vector Form of Ohm's Law

The vector form of Ohm's Law states that for an isotropic conductor (a material whose properties are the same in all directions), the current density ($\vec{J}$) at any point is directly proportional to the electric field ($\vec{E}$) at that point. The constant of proportionality is the electrical conductivity ($\sigma$) of the material.

$ \vec{J} = \sigma \vec{E} $

This equation tells us that the direction of the current density vector ($\vec{J}$, the direction of conventional current flow) is the same as the direction of the electric field vector ($\vec{E}$) within the conductor. The magnitude of the current density is proportional to the magnitude of the electric field, with conductivity $\sigma$ as the proportionality constant.

Using the resistivity $\rho = 1/\sigma$, the vector form can also be written as:

$ \vec{E} = \rho \vec{J} $

This form shows that the electric field required to produce a certain current density is proportional to the resistivity of the material.

---

Derivation of Scalar Ohm's Law ($V=IR$) from Vector Form ($\vec{J} = \sigma \vec{E}$)

We can derive the familiar scalar form $V=IR$ from the vector form $\vec{J} = \sigma \vec{E}$ for a uniform conductor.

Consider a cylindrical conductor of length $L$, uniform cross-sectional area $A$, and uniform conductivity $\sigma$. Let a potential difference $V$ be applied across its ends, creating a uniform electric field $\vec{E}$ inside, parallel to the length of the conductor.

The relationship between the magnitude of the electric field $E$ and the potential difference $V$ across the length $L$ is $V = E L$. So, $E = V/L$.

The current $I$ flows uniformly through the cross-section $A$. The magnitude of the current density is $J = I/A$. The direction of $\vec{J}$ is the same as $\vec{E}$.

From the magnitude of the vector form of Ohm's Law, $J = \sigma E$.

Substitute the expressions for $J$ and $E$:

$ \frac{I}{A} = \sigma \left(\frac{V}{L}\right) $

Rearranging this equation to solve for $V$:

$ V = I \left(\frac{L}{\sigma A}\right) $

We know that the resistance $R$ of a uniform conductor is given by $R = \frac{L}{\sigma A}$ (since $\sigma = 1/\rho$, $R = \rho L/A$).

Substituting this into the equation for $V$:

$ V = I R $

This successfully derives the scalar form of Ohm's Law for a uniform conductor from its more general vector form. The vector form $\vec{J} = \sigma \vec{E}$ is more fundamental as it applies at every point within the conductor, even if the conductor is not uniform or the electric field is not constant throughout.

---

Anisotropic Materials

For certain materials, known as anisotropic materials (e.g., some crystals), the electrical properties are direction-dependent. In such materials, the current density $\vec{J}$ may not be in the same direction as the electric field $\vec{E}$. In this case, conductivity $\sigma$ is not a simple scalar but a tensor (a mathematical entity representing properties that vary with direction). The vector form of Ohm's Law is then expressed using matrix notation for the conductivity tensor. However, for most common conductors (like metals and semiconductors) treated as isotropic, $\sigma$ is a scalar.